# Examples of path integral where path of extremal action does not contribute the most?

I have learnt that by doing a saddle point approximation in the path integral formulation of quantum mechanics, the classical action (extremal action where $$\delta S=0$$) is the one that contributes the most, hence seeing how classical physics arises from quantum physics.

The question is: is there any example in quantum physics (especially QFT) where the most contribution does not come from the path of extremal action? I know that the proof using the saddle point approximation seems very general, but I was thinking that there might be some peculiar/interesting terms in the action (such as topological) where the path with $$\delta S=0$$ do not contribute the most?

• To clarify: The contribution of any specific path is zero, whether or not it satisfies $\delta S=0$. The saddle-point approximation works by considering a neighborhood of a path with $\delta S=0$, and it's the "volume" of the neighborhood that makes the contribution non-zero. (More precisely, we expand $S$ to second order about the saddle point and do the resulting Gaussian integral.) Is the question implicitly asking if there are situations where a similar quadratic expansion about some other classical configuration gives a better approximation? May 24, 2020 at 1:21
• @ChiralAnomaly Yes, I believe you can pose the question in this way. Thanks. May 24, 2020 at 11:23
• An obvious example where the trajectory of extremal action doesn't contribute at all is quantum tunelling, where there isn't a classically allowed trajectory at all. See the references [34–38] in this paper.
– m93a
Apr 7, 2021 at 13:03

In general it seems quite unlikely, unless there is some special symmetry that enforces this. I guess it could also sometimes happen in 0+1 dimensions because here the space of fields is much nicer.

But anyway, the closest example that I can think of is $$\mathcal N=2$$ supersymmetry in four dimensions. Here, due to supersymmetry, the effective action is one-loop exact, and the only corrections are non-perturbative: $$F=\Psi^2\log\Psi^2+\text{instantons}$$ where the instantons come from classical field configurations with a macroscopic value. These have non-vanishing action $$S\sim 1/g^2$$, so their contribution is $$e^{-1/g^2}$$ and thus very small.

Anyway, at zero scalar field the classical equations of motion admit non-trivial instanton solutions (cf ref. 1), $$A\sim \frac{1}{x^2+\rho^2},\qquad \psi\sim\frac{\gamma}{x^2+\rho^2}$$ where $$\rho$$ is the size of the instanton.

For non-zero scalar field, this is no longer a solution to the classical equations of motion, but thanks to supersymmetry one can argue that they still yield the leading contribution to $$F$$ (essentially, because other contributions vanish!).

So this is a situation where field configurations that do not strictly speaking satisfy the classical equations of motion still give the largest contribution to the path integral. The reason is, other contributions are exactly zero due to the large amount of symmetry provided by $$\mathcal N=2$$. This is a quite non-generic situation, and you would not expect it to happen in more realistic theories.

References.

• Aren't the instantons also satisfying the Euclidean eom, so that they are actually also saddle-points of the action? Apr 13, 2021 at 12:31
• @ohneVal Yes, instantons by definition satisfy the eom. Not sure what your objection is. As I said above, the expressions for $A\psi$ are instantons (they satisfy the eom) only if the scalar vev vanishes. For non-vanishing scalar they are not instantons (they do not satisfy the eom). Still, despite not being actual instantons, they give the largest contribution to the superpotential. Apr 14, 2021 at 15:37
• I missed the part about the vev, but I am interested in understanding that better, how to they vanish? Do you mean there is no more instanton solutions in this supersymmetric case? Apr 14, 2021 at 20:24

The answer by @Accidental FourierTransform has already mentioned the instantons, but let me give a more pedestrian example and a more pedestrian view on the question.

When we use a path integral to describe the motion of a single quantum particle, the saddle point approximation corresponds to what is otherwise known as quasiclassical approximation, and the justification for it is literally taking $$\hbar\rightarrow 0$$. (As a trivia: Landau & Livshitz in their QM book literally derive quasiclassical approximation from what they call action using eikonal approximation without ever referring to path integrals.)

This provides us with a simple hint of where the saddle point approximation may not work: where quasiclassical approximation doesn't. A well-known example is the tunneling between two degenerate potential minima, where the correct splitting of the energy levels can be recovered via the alraedy mentioned instantons (which can be also viewed as a saddle-point approximation after transformation to the Eucledean space.)